The molar mass of gallium (Ga) is 69.72 g/mol.
Calculate the number of atoms in a 27.2 mg sample of Ga.
Write your answer in scientific notation using three significant figures.
atoms Ga
Answer:
2.35 x 10²⁰ atoms Ga
Explanation:
After converting from mg to g, use the molar mass as the unit converter to convert to moles. Then using Avogadro's number, 6.022 x 10²³ convert from moles to atoms of Ga.
[tex]27.2mgGa*\frac{1g}{1000mg} *\frac{1 mol Ga}{69.72gGa} *\frac{6.022*10^2^3 atoms Ga}{1 molGa} = 2.349 * 10^2^0 atoms Ga[/tex]
Then round to 3 significant figures = 2.35 x 10²⁰ atoms Ga.
The number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
StoichiometryFrom the question, we are to calculate the number of atoms in a 27.2 mg sample of Ga.
First, we will determine the number of moles of Ga present
Using the formula,
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass} [/tex]
Mass = 27.2 mg = 0.0272 g
Molar mass = 69.72 g/mol
Then,
[tex]Number\ of\ moles \ of\ Ga = \frac{0.0272}{69.72} [/tex]
[tex]Number\ of\ moles \ of\ Ga = [/tex] 0.000390132 moles
Now, for the number of atoms present
From the formula
Number of atoms = Number of moles × Avogadro's constant
Then,
Number of Ga atoms = 0.000390132 × 6.022×10²³
Number of Ga atoms = 2.35 × 10²⁰ atoms
Hence, the number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
Learn more on stoichiometry here: https://brainly.com/question/14464650
What is the pH of a solution made by mixing 0.050 mol of NaCN with enough water to make a liter of solution
Answer:
pH = 11
Explanation:
The equilibrium of a weak base as NaCN in water is:
NaCN(aq) + H₂O(l) ⇄ OH⁻(aq) + Na⁺(aq) + HCN(aq)
And kb, the equilibrium constant, is:
Kb = [OH⁻] [HCN] / [NaCN]
Where Kb of NaCN is 2.04x10⁻⁵
In the beginning, the [NaCN] is 0.050mol / L = 0.050M.
Both [OH⁻] and [HCN] are produced from this equilibrium, and its concentration is X, that is:
2.04x10⁻⁵ = [X] [X] / [0.050M]
1.02x10⁻⁶ = X²
X = 1x10⁻³ = [OH⁻]
As pOH = - log [OH⁻]
pOH = 3.00
And pH = 14 - pOH
pH = 11
plz help answer both will mark brainest
A balloon contains 1.1 L of gas at a pressure of 0.80 atm. How will the volume
change if the pressure is increased to 2.0 atm?
Answer:
Final volume = 0.44 L
Explanation:
Given data:
Initial volume of balloon = 1.1 L
Initial pressure = 0.80 atm
Final volume = ?
Final pressure = 2.0 atm
Solution:
The given problem will be solved through the Boly's law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
0.80 atm × 1.1 L = 2.0 atm × V₂
V₂ = 0.88 atm. L/ 2.0 atm
V₂ = 0.44 L
What is the most highly populated rotational level of Cl2 (i) 25deg C and (ii) 100 deg C? Take B=0.244cm-1.This question should not be resubmitted, it is a textbook question from the Atkins physical chemistry txtbook. 10 e.
Answer:
i
[tex]J_{m} = 20 [/tex]
ii
[tex]J_{m} = 22.5 [/tex]
Explanation:
From the question we are told that
The first temperatures is [tex]T_1 = 25^oC = 25 +273 =298 \ K[/tex]
The second temperature is [tex]T_2 = 100^oC = 100 +273 = 373 \ K[/tex]
Generally the equation for the most highly populated rotational energy level is mathematically represented as
[tex]J_{m} = [ \frac{RT}{2B}] ^{\frac{1}{2} } - \frac{1}{2}[/tex]
Here R is the gas constant with value [tex]R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}[/tex]
Also
B is given as [tex]B=\ 0.244 \ cm^{-1}[/tex]
Generally the energy require per mole to move 1 cm is 12 J /mole
So [tex]0.244 \ cm^{-1}[/tex] will require x J/mole
[tex]x = 0.244 * 12[/tex]
=> [tex]x = 2.928 \ J/mol [/tex]
So at the first temperature
[tex]J_{m} = [ \frac{8.314 * 298 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 20 [/tex]
So at the second temperature
[tex]J_{m} = [ \frac{8.314 * 373 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 22.5 [/tex]
Josh heated a certain amount of blue copper sulfate crystals to get 2.1 g of white copper sulfate powder and 1.4 g of water. What is most likely the mass of the blue copper sulfate that he heated and why?
Answer: The mass of blue copper sulfate is 3.5 g
Explanation:
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
The chemical equation for the heating of copper sulfate crystals is:
Let the mass of blue copper sulfate be 'x' grams
We are given:
Mass of copper sulfate powder = 2.1 grams
Mass of water = 1.4 grams
Total mass on reactant side = x
Total mass on product side = (2.1 + 1.4) g
So, by applying law of conservation of mass, we get:
Hence, the mass of blue copper sulfate is 3.5 grams
If the earth was a guava fruit, the space where the seeds are would be the core/mantle
1. What 2 subatomic particles have charges? List the particle name and its charge.
Answer: Proton - positive charge (+)
Neutron - neutral charge (0)
Electron - negative charge (-)
Explanation:
PLEASE HELP! WILL DO BRAINLIEST! What do scientists call all of the compounds that contain carbon and are found in living things?
organic
inorganic
acidic
nonacidic
Answer:
acidic because of electrical issues and the body of electrical equipment
Which is one way that minerals crystallize from materials dissolved in water?
from the air
from solutions that evaporate
from hot water solutions when water boils
from the soil
Answer:
the second answer its science behind it
Answer:
b
Explanation:
2 2 6 2 6 2 10 3
1s 2s 2p 3s 3p 4s 3d 4p
=
Answer:
ARSENIC
Explanation:
It has an atomic number of 33
A student measured the masses of four different-sized blocks. The student determined that each block had a mass of 50 grams.
(There is a small block, a little bit bigger block, a big block and the biggles block)
Which block has the least density?
Answer:..
Explanation:
In the laboratory you dissolve 18.7 g of copper(II) bromide in a volumetric flask and add water to a total volume of 375mL.
Required:
a. What is the molarity of the solution?
b. What is the concentration of the copper(II) cation?
c. What is the concentration of the acetate anion?
Answer:
a) - 0.2 M
b) - 0.2 M
c)- 0
Explanation:
The chemical formula of copper (II) bromide is CuBr₂. Its molar mass (MM) is calculated as follows:
MM(CuBr₂)= MM(Cu) + (2 x MM(Br) = 63.5 g/mol + (2 x 80 g/mol)= 223.5 g/mol
a). Molarity = moles CuBr₂/1 L solution
moles CuBr₂ = mass/MM = 18.7 g x 1 mol/223.5 g = 0.084 mol
Volume in L = 375 mL x 1 L/1000 mL = 0.375 L
M = 0.084 mol/(0.375 L) = 0.223 M ≅ 0.2 M
b). When is added to water, CuBr₂ dissociates into ions as follows:
CuBr₂ ⇒ Cu²⁺ + 2 Br⁻
We have 1 mol Cu²⁺ (copper (II) cation) per mol of CuBr₂. Thus, the concentration of copper (II) cation is:
0.2 mol CuBr₂ x 1 mol Cu²⁺/mol CuBr₂ = 0.2 M
c). The concentration of acetate anion is 0. There is no acetate anion in the solution (the anion from CuBr₂ is bromide Br⁻).
SOMEONE PLZ HELP!!!!
Answer:
4.22mL
Explanation:
V=m/d
v= 18.45g/4.37g/mL
Problem:
[Ar]4s2
Identify the period (p) , group (g) and valence electrons block of the element
Answer:
it is Calcium (Ca)
4th period, 2nd group, 2 valence electrons
Solid diarsenic trioxide reacts with fluorine gas (F2) to produce liquid arsenic pentafluoride and oxygen gas (O2). Write the Qc for this reaction.
Answer:
QC= [O2]^3/[F2]^10
Explanation:
If you collect 5.74 mL of O 2 at 298 K and 1.00 atm over 60.0 seconds from a reaction solution of 5.08 mL, what is the initial rate of the reaction
Answer:
7.71 × 10⁻⁴ M/s
Explanation:
The initial rate of the reaction can be expressed by using the formula:
[tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]
where the number of moles of O₂ = [tex]\dfrac{PV}{RT}[/tex]
where;
Pressue P = 1.00 atm
Volume V =5.74mL = (5.74 /1000) L
Rate R = 0.082 L atm/mol.K
Temperature = 298 K
[tex]= \dfrac{1.00 \ atm \times \dfrac{5.74 }{1000}L}{0.082 \ L \ atm/mol.K \times 298 K}[/tex]
= 2.35 × 10⁻⁴ mol
Δ[O₂] = [tex]\dfrac{moles \ produced - initial \ mole}{\dfrac{5.08 }{1000}L }[/tex]
Δ[O₂] = [tex]\dfrac{2.35 \times 10^{-4} M - 0 M}{\dfrac{5.08 }{1000}}[/tex]
Δ[O₂] = 0.04626 M
The initial rate = [tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]
= [tex]\dfrac{0.04626}{60}[/tex]
= 7.71 × 10⁻⁴ M/s